Question 1

An ideal incompressible fluid flows at 12 m/s in a horizontal pipe.If the pipe widens to twice its original radius,what is the flow speed in the wider section?

Accepted Answer

According to the principle of continuity for incompressible fluids, the flow rate must remain constant. If the radius doubles, the cross-sectional area increases by a factor of four (since area is proportional to the square of the radius). Therefore, the flow speed must decrease by a factor of four to maintain the same flow rate, resulting in a speed of 3.0 m/s.

Question 2

Water flows through a horizontal pipe of cross-sectional area 10.0 cm² at a pressure of 0.250 atm with a flow rate is 1.00 L/s.At a valve,the effective cross-sectional area of the pipe is reduced to 5.00 cm².What is the pressure at the valve? The density of water is 1000 kg/m³,and treat it as an ideal incompressible fluid.

Accepted Answer

We can use the principle of continuity to find the velocity of water before and after the valve:
A1v1 = A2v2
where A1 = 10.0 cm^2 = 0.001 m^2, A2 = 5.00 cm^2 = 0.0005 m^2, v1 = Q/A1 = 0.1 m/s, and Q = 1.00 L/s = 0.001 m^3/s.
So, v2 = (A1/A2)v1 = 0.2 m/s.

Next, we can use Bernoulli's equation to relate the pressure at the two points:
P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
where P1 = 0.250 atm = 25398.8 Pa and ρ = 1000 kg/m^3.
Solving for P2, we get:
P2 = P1 + 1/2ρ(v1^2 - v2^2) = 0.235 atm = 23912.9 Pa.
Therefore, the pressure at the valve is B) 0.235 atm.

Question 3

Glycerin,with a viscosity of 1.5 N ∙ s/m²,is flowing with an average speed of 1.2 m/s through a tube that has a radius of 5.0 mm and is 25 cm long.What is the drop in pressure over the length of this pipe?

Accepted Answer

The drop in pressure is given by the Hagen-Poiseuille equation:$$\Delta P = \frac{8\mu L Q}{\pi r^4}$$where:* $\Delta P$ is the drop in pressure (Pa)* $\mu$ is the viscosity (N ∙ s/m²)* $L$ is the length of the pipe (m)* $Q$ is the flow rate (m³/s)* $r$ is the radius of the pipe (m)We can calculate the flow rate from the average speed and the cross-sectional area of the pipe:$$Q = Av = \pi r^2 v$$where:* $A$ is the cross-sectional area of the pipe (m²)* $v$ is the average speed (m/s)Substituting the given values into the Hagen-Poiseuille equation, we get:$$\Delta P = \frac{8\mu L Q}{\pi r^4} = \frac{8(1.5 ext{ N} \cdot ext{s/m}^2)(0.25 ext{ m})(\pi (5.0 imes 10^{-3} ext{ m})^2 (1.2 ext{ m/s}))}{\pi (5.0 imes 10^{-3} ext{ m})^4} = 140 ext{ kPa}$$Therefore, the drop in pressure over the length of the pipe is 140 kPa.

Question 4

Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm.What is the volume flow rate of the fluid?

Accepted Answer

The volume flow rate (Q) of a fluid is given by Q = A × v, where A is the cross-sectional area of the pipe and v is the velocity of the fluid.

The diameter of the pipe is given as 3.0 cm, which means the radius (r) is 1.5 cm (or 0.015 m). The cross-sectional area of the pipe can be calculated as:

A = πr^2 = π × (0.015 m)^2 = 7.07 × 10^-7 m^2

The velocity of the fluid is given as 2.0 m/s. Using these values, we can calculate the volume flow rate as:

Q = A × v = 7.07 × 10^-7 m^2 × 2.0 m/s = 1.4 × 10^-3 m^3/s

Therefore, the best choice is (A) 1.4 × 10^-3 m^3/s.

Question 5

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous value.If the speed of the water in the larger section of the pipe was 
What is its speed in this smaller section if the water behaves like an ideal incompressible fluid?

Accepted Answer

The answer of Water flowing through a cylindrical pipe suddenly...

Question 6

Consider a very small hole in the bottom of a tank that is 
In diameter and filled with water to a height of 
Find the speed at which the water exits the tank through the hole.

Accepted Answer

The answer of Consider a very small hole in the...

Question 7

In a section of horizontal pipe with a diameter of 3.0 cm,the pressure is 100 kPa and water is flowing with a speed of 1.5 m/s.The pipe narrows to 2.0 cm.What is the pressure in the narrower region? Treat the water as an ideal incompressible fluid.

Accepted Answer

To find the pressure in the narrower region, we can use the Bernoulli's equation, which in one of its forms can be written as $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P_1$ and $P_2$ are the pressures at points 1 and 2, $v_1$ and $v_2$ are the velocities at points 1 and 2, and $\rho$ is the density of the fluid (water in this case). The equation assumes the fluid is incompressible and flow is steady.Given:- $P_1 = 100$ kPa- $v_1 = 1.5$ m/s- Diameter at point 1, $D_1 = 3.0$ cm or 0.03 m- Diameter at point 2, $D_2 = 2.0$ cm or 0.02 mFirst, we find the velocities using the continuity equation $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas at points 1 and 2, respectively. The area of a circle is given by $A = \pi r^2$, so we can find $v_2$ as follows:$$A_1 = \pi (0.015)^2$$$$A_2 = \pi (0.01)^2$$$$v_2 = v_1 \times \frac{A_1}{A_2} = 1.5 \times \frac{\pi (0.015)^2}{\pi (0.01)^2} = 1.5 \times \frac{(0.015)^2}{(0.01)^2} = 1.5 \times 2.25 = 3.375 \, \text{m/s}$$Now, applying Bernoulli's equation to solve for $P_2$:$$100 \times 10^3 + \frac{1}{2} \times 1000 \times (1.5)^2 = P_2 + \frac{1}{2} \times 1000 \times (3.375)^2$$Solving for $P_2$:$$P_2 = 100 \times 10^3 + \frac{1}{2} \times 1000 \times (1.5)^2 - \frac{1}{2} \times 1000 \times (3.375)^2$$$$P_2 = 100000 + 1125 - 5681.25 = 100000 - 4556.25 = 95443.75 \, \text{Pa} = 95.44 \, \text{kPa}$$Rounding to the nearest kPa gives us 95 kPa, which corresponds to choice A.

Question 8

A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a fluid through a pipe.What is the speed of the fluid when it enters a portion of the pipe that is 3.0 mm in diameter? Treat the fluid as ideal and incompressible.

Accepted Answer

The speed of the fluid increases as it moves from a wider to a narrower section of the pipe due to the conservation of mass principle for an incompressible fluid, which can be described by the equation $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas of the wider and narrower sections, respectively, and $v_1$ and $v_2$ are the fluid velocities in these sections. Given the diameters, the areas can be calculated using $A = \pi r^2$, leading to a significant increase in velocity when the fluid moves into the narrower section. Calculating with the given numbers, the velocity in the narrower section is found to be 50 m/s.

Question 9

Ideal incompressible water flows through a horizontal pipe of cross-sectional area 10.0 cm² at a pressure of 0.250 atm with a volume flow rate of 1.00 × 10^-3 m³/s.At a valve,the effective cross-sectional area of the pipe is reduced to 5.00 cm².What is the pressure at the valve?

Accepted Answer

The volume flow rate is conserved in an incompressible fluid, so we have:

$A_1 v_1 = A_2 v_2$

where $A_1$ and $v_1$ are the cross-sectional area and velocity before the valve, and $A_2$ and $v_2$ are the cross-sectional area and velocity after the valve. We can rearrange this equation to solve for $v_2$:

$v_2 = \frac{A_1}{A_2} v_1$

We also know that the fluid is incompressible, so the mass flow rate is conserved:

$ho_1 A_1 v_1 = ho_2 A_2 v_2$

where $ho_1$ and $ho_2$ are the densities before and after the valve. We can rearrange this equation to solve for $ho_2$:

$ho_2 = \frac{ho_1 A_1 v_1}{A_2 v_2}$

Since the density of water is (approximately) constant, we can use this equation to find the pressure after the valve. The pressure difference across the valve is given by:

$\Delta P = ho_2 g h$

where $g$ is the acceleration due to gravity and $h$ is the difference in height between the two points. In this case, the two points are at the same height, so $\Delta P = 0$. Therefore we have:

$P_1 + \frac{1}{2} ho_1 v_1^2 = P_2 + \frac{1}{2} ho_2 v_2^2$

Solving for $P_2$ and substituting in the given values, we get:

$P_2 = P_1 + \frac{1}{2} (ho_1 v_1^2 - ho_2 v_2^2)$

$P_2 = 0.250 \, 	ext{atm} + \frac{1}{2} \left[\left(1000 \frac{	ext{kg}}{	ext{m}^3}ight) \left(10^{-2} \, 	ext{m}^2ight) \left(1.00 	imes 10^{-3} \frac{	ext{m}}{	ext{s}}ight)^2 - \left(1000 \frac{	ext{kg}}{	ext{m}^3}ight) \left(5.00 	imes 10^{-4} \, 	ext{m}^2ight) \left(\frac{10^{-2} \, 	ext{m}}{2.00 	imes 10^{-3} \, 	ext{m}/	ext{s}}ight)^2ight]$

$P_2 = 0.235 \, 	ext{atm}$

Therefore the answer is D.

Question 10

Ideal incompressible water is flowing in a drainage channel of rectangular cross-section.At one point,the width of the channel is 12 m,the depth of water is 6.0 m,and the speed of the flow is 2.5 m/s.At a point downstream,the width has narrowed to 9.0 m,and the depth of water is 8.0 m.What is the speed of the flow at the second point?

Accepted Answer

The answer of Ideal incompressible water is flowing in a...

Question 11

For a certain patient,the build-up of fatty tissue on the wall of an artery has decreased the arterial radius by 10%.By how much would the pressure provided by the heart have to be increased to maintain a constant volume of blood flow? Model the blood as an ideal incompressible fluid.

Accepted Answer

The answer of For a certain patient,the build-up of fatty...

Question 12

Water is flowing in a drainage channel of rectangular cross-section.The width of the channel is 15 m,the depth of water is 8.0 m,and the speed of the flow is 2.5 m/s.What is the mass flow rate of the water? The density of water is 1000 kg/m³.

Accepted Answer

The mass flow rate is calculated by multiplying the cross-sectional area of the channel, the flow speed, and the density of water. The area is 15 m * 8 m = 120 m². The flow rate is 120 m² * 2.5 m/s = 300 m³/s. The mass flow rate is 300 m³/s * 1000 kg/m³ = 3.0 × 10⁵ kg/s.

Question 13

Water flows out of a large reservoir through an open pipe,as shown in the figure.What is the speed of the water as it comes out of the pipe?

Accepted Answer

Using the principle of conservation of energy, we can equate the potential energy of the water at the top of the reservoir to the kinetic energy of the water at the bottom of the pipe.

mgh = 1/2 mv^2

where m is the mass of the water, g is the acceleration due to gravity, h is the height of the reservoir, and v is the velocity of the water at the bottom of the pipe.

Solving for v, we get:

v = sqrt(2gh)

Substituting the given values, we get:

v = sqrt(2 × 9.81 m/s^2 × 40 m) ≈ 13 m/s

Therefore, the speed of the water as it comes out of the pipe is approximately 13 m/s. Hence, the correct choice is C).

Question 14

Water,which we can treat as ideal and incompressible,flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10⁴ Pa.If the pipe widens to twice its original radius,what is the pressure in the wider section?

Accepted Answer

The answer of Water,which we can treat as ideal and...

Question 15

An incompressible ideal fluid flows steadily through a pipe that has a change in diameter.The fluid speed at a location where the pipe diameter is 8.0 cm is 1.28 m/s.What is the fluid speed at a location where the diameter has narrowed to 4.0 cm?

Accepted Answer

The answer of An incompressible ideal fluid flows steadily through...

Question 16

A hole of radius 1.0 mm forms in the bottom of a very large water storage tank that holds water to a depth of 15 m.At what volume rate will water flow out of the hole?

Accepted Answer

The answer of A hole of radius 1.0 mm forms...

Question 17

A pressurized cylindrical tank,5.0 m in diameter,contains water that emerges from the pipe at point C with a speed of As shown in the figure.Point A is 10 m above point B and point C is 3.0 m above point B.The area of the pipe at point B is And the pipe narrows to an area of At point C.Assume that the water is an ideal fluid in laminar flow.The density of water is 1000 kg/m³ The mass flow rate in the pipe is closest to

Accepted Answer

The answer of A pressurized cylindrical tank,5.0 m in diameter,contains...

Question 18

Ideal incompressible fluid flows through a 4.0-cm-diameter pipe at 1.0 m/s.There is a 2.0 cm diameter constriction in the line.What is the speed of the fluid in this constriction?

Accepted Answer

According to the principle of continuity for incompressible fluids, the product of cross-sectional area and velocity is constant. The area of the larger pipe is π(2.0 cm)^2 and the area of the constriction is π(1.0 cm)^2. The velocity in the constriction is (4.0 cm/2.0 cm)^2 * 1.0 m/s = 4.0 m/s.

Question 19

Water flows out of a large reservoir through 5.0-cm diameter pipe.The pipe connects to a 3.0-cm diameter pipe that is open to the atmosphere,as shown in the figure.What is the speed of the water in the 5.0-cm pipe? Treat the water as an ideal incompressible fluid.

Accepted Answer

The answer of Water flows out of a large reservoir...

Question 20

A pressurized cylindrical tank,5.0 m in diameter,contains water that emerges from the pipe at point C with a speed of As shown in the figure.Point A is 10 m above point B and point C is 3.0 m above point B.The area of the pipe at point B is And the pipe narrows to an area of At point C.Assume that the water is an ideal fluid in laminar flow.The density of water is 1000 kg/m³.The rate at which the water level is falling in the tank is closest to

Accepted Answer

The answer of A pressurized cylindrical tank,5.0 m in diameter,contains...

Quiz 15: Fluids